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Consider the following Hamiltonian: $$\hat H=\frac{\hbar\omega}{2}(\hat x^2+\hat p^2)-\frac{\hbar\omega}{2}\hat 1 =\frac{\hbar\omega}{2}(\hat x^2+\hat p^2-\hat 1 )$$

After defining annihilation and creation operators in the usual way, this can be written as:$$\hat H=\left(\hat a^{\dagger}\hat a +\frac{\hat 1}{2}-\frac{\hat 1}{2}\right)\hbar\omega$$ Thus,

$$E_n=n\hbar\omega$$

Now, in the case of a harmonic oscillator we know from the normalization condition of number states that $n\geq0$ is satisfied and hence $n$ can be taken zero for the ground state. For a HO this gives the ground state energy as $\hbar\omega/2$ but here it gives zero. I tried to see if there was something that constraints $n$ to be greater than $0$ but I couldn't find it.

Also, since the ground state for this system can be built explicitly from the definitions of $\hat a$ and $\hat a ^{\dagger}$ so I think that the ground wave function for this case will also not go to zero if $n=0$. Therefore, this cannot be a reason to constraint $n$ to be greater than zero.

Does that mean that ground state energy in this case can be zero?

But this conclusion seems wrong to me since I have just added a constant factor to the Hamiltonian of a Harmonic oscillator which is nothing but shifting the potential level and I don't think that should change the zero-point energy.

So, what's going on here?

EDIT:

Maybe I have not indicated my doubt clearly.

I know that a constant offset in the hamiltonian does not change the dynamics of the system.

I also understand that for many systems we generally make the ground state energy as zero reference assigning it a value of zero as we are interested only in the excitations or difference in energy levels.

The problem that I have is that I think (I might be wrong) that zero-point energy represents real fluctuations arising due to the uncertainty principle and this can be seen in a $0$K scenario where even at that temperature there is still kinetic energy.

But as can be seen in the above example, this energy seems to come out to be zero. What does this mean? How is it consistent with the fact that the lowest state should have some energy?

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    $\begingroup$ "since I have just added a constant factor to the Hamiltonian of a Harmonic oscillator which is nothing but shifting the potential level and I don't think that should change the zero point energy" - what do you think the "potential level" is if not the energy? $\endgroup$ Commented Aug 4, 2021 at 17:17
  • $\begingroup$ You might find this question, and the associated answers interesting $\endgroup$ Commented Aug 4, 2021 at 18:32
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    $\begingroup$ @ACuriousMind To put my question in a better context let me quote the part from an answer given here by Sahand: physics.stackexchange.com/q/614186 $\endgroup$ Commented Aug 5, 2021 at 9:18
  • $\begingroup$ "I think what is typically meant by "zero-point energy" in this context is actually zero-point kinetic energy. Exactly as you say, a constant offset $E_0$ in the Hamiltonian $H=H_0+E_{0}$ has no consequences on the dynamics and classically speaking, it corresponds to picking a different reference point for your potential energy function." $\endgroup$ Commented Aug 5, 2021 at 9:19
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    $\begingroup$ Please edit information that clarifies your question into the question instead of just leaving it in comments $\endgroup$ Commented Aug 7, 2021 at 11:21

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I'll start with things you already understand. The question considers a model with Hamiltonian $$ H \propto x^2+p^2-1 \tag{1} $$ with a positive overall factor, where $x$ and $p$ are operators satisfying $$ [x,p]=i. \tag{2} $$ As noted in the question, the Hamiltonian can also be written $$ H\propto a^\dagger a \tag{3} $$ with $$ a\propto x-ip. \tag{4} $$ The eigenstate of $H$ with the lowest eigenvalue is the state $|0\rangle$ that satisfies $$ a|0\rangle = 0. \tag{5} $$ This is the ground state. The eigenvalue (which is sometimes called "zero point energy") is irrelevant, because shifting $H$ by a constant term does not change the fact that (5) has the lowest eigenvalue among all eigenstates of $H$. You already understand this.

I'm not sure what you want to call "kinetic energy" in this context, but suppose it's the operator $$ K \propto p^2. \tag{6} $$ The ground state an eigenstate of the total energy operator $H$, so measuring $H$ in the ground state would always give the same outcome. But the ground state is not an eigenstate of the kinetic energy operator $K$. If we measure $K$ in the state $|0\rangle$, and if we repeat this experiment a jillion times, we'll get a jillion different outcomes. This is an example of the so-called uncertainty principle: if you measure an observable in a state that is not one of the observable's eigenstates, you get diverse outcomes. Sometimes people describe this using words like "vacuum fluctuations," but I don't know why. Maybe they're imagining Hidden Variables.

When we say that a system is at absolute zero, we mean that it is in its ground state — the state of lowest total energy. In this example, that's the state satisfying (5). In some systems, like an ideal gas, total energy and kinetic energy are the same thing. The harmonic oscillator is not one of those systems, at least not if "kinetic energy" means (6). In general, zero absolute temperature does not mean zero kinetic energy. Zero absolute temperature does mean time-independent, and the ground state — the ray represented by the vector $|0\rangle$ — is indeed independent of time.

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  • $\begingroup$ Thanks. A very illuminating answer. So for a harmonic oscillator the toal energy in the ground state can be zero while the kinetic energy cannot be? $\endgroup$ Commented Aug 7, 2021 at 21:06
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    $\begingroup$ @Lost I would say it this way: for a harmonic oscillator, the total energy in the ground state is zero (if we choose the constant term as you did), but the kinetic energy does not have any well-defined value, because $|0\rangle$ is not an eigenstate of $K$. The kinetic energy does have a well-defined expectation value $\langle 0|K|0\rangle$, which is positive (not zero), but that doesn't mean anything is moving, not even stochastically. The ground state is a stationary state. $\endgroup$ Commented Aug 7, 2021 at 21:14
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    $\begingroup$ @Lost "Non-zero kinetic energy but no motion!?" That would be a contradiction in classical physics, but this is quantum physics. Eigenstates of the Hamiltonian (I mean the corresponding rays) are time-independent. The ground state is just one example. To get motion, and especially to get anything approximating an eigenstate of the kinetic energy $K$, you need a superposition of different eigenstates of the Hamiltonian with different eigenvalues (different total energies). $\endgroup$ Commented Aug 7, 2021 at 22:43
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    $\begingroup$ @Lost I found a copy of the book (2nd edition), and I found the sentence you quoted. The book defines zero-point energy to be $\hbar^2\pi^2/(2ma^2)$ for the square well and to be $\hbar\omega/2$ for the harmonic oscillator. Then it goes on to make the quoted assertion about the zero-point energy of all bound state systems, but it doesn't give a general definition of zero-point energy, so that assertion is meaningless. Not even wrong, just plain meaningless. The rest of the book still looks useful, though, so don't throw it away. No physics book is perfect. They're all written by humans. $\endgroup$ Commented Aug 8, 2021 at 1:40
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    $\begingroup$ @Lost By the way, the bit about kinetic energy is mentioned in problem 4.2 (in the 2nd edition), except that it's worded in a way that might be misleading. The book says the kinetic energy is nonzero. It should say that the kinetic energy is not equal to zero -- or to any other individual real number. $\endgroup$ Commented Aug 8, 2021 at 1:40
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For any Hamiltonian $H$ with zero-point energy $E_0$, the shifted Hamiltonian $$ H'=H-E_0 $$ has vanishing zero-point energy, by construction.

The zero-point energy is unphysical, you can redefine it away at will. Your own argument shows this. Your argument is ok. There is nothing constraining $n$. The Hamiltonian you constructed has no zero-point energy.

The zero-point energy does not represent "fluctuations". There is no such thing as "fluctuations". All there is, is quantum mechanics. You have a quantum system, and the system has properties. There is no point in calling these properties "fluctuations".

The zero-point energy is unphysical and undetectable. What is physical, what has real world consequences, is differences in energies. Physical phenomena that we usually ascribe to a zero-point energies are actually nothing but the effect in the difference between two energies, not an effect due to a single energy being zero or not.

For example, the Casimir effect is not due to the zero-point energy. It is due to the difference in energy with and without the plaque. (Or more precisely, the zero of energies is defined to be at $L\to\infty$, if you will.)

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  • $\begingroup$ Okay. But take a simple harmonic oscillator and keep on removing energy units from it. You will have a stage where you can't take out more energy. The energy that's left is the difference between what ? More precisely between which energy levels? We only have one energy whose value we can't make zero. Wouldn't it be weird to simply say this is the difference of energy between the presence and absence of the harmonic oscillator? Also, in this context I've read about zero-point kinetic energy and getting zero-point expectation values of $x$ and $p$. $\endgroup$ Commented Aug 7, 2021 at 20:24
  • $\begingroup$ Could you please elaborate whether these things are related ,if at all, to zero point energy? (Note in this question, it is implicit that I am also asking how zero point energy is related to heisenberg's uncertainty principle) $\endgroup$ Commented Aug 7, 2021 at 20:26
  • $\begingroup$ Zero-point energy is not unphysical, see Why not drop ℏω/2 from the quantum harmonic oscillator energy? The accepted answer (mine) there gives an example when it's detectable. $\endgroup$ Commented Aug 16, 2021 at 13:45
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The zero point energy is a consequence of the wave nature of quantum states. The ground state of a harmonic oscillator has a position and momentum uncertainty. You can understand the momentum uncertainty in a way that the state does "never rest". It is always a superposition of momenta, thus you get the energy. If you add another constant, you have to do this to the classical analogue too. The classical analogue is then allowed to have a negative energy. You would then redefine this as a new zero, and measured from this point, you would then get a new zero point energy in the quantum version.

This is a genuine concern. The behaviour of some people on here in the comments is outright shameful.

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  • $\begingroup$ Thank you a lot for acknowledging that this was a genuine question that one could face! $\endgroup$ Commented Aug 7, 2021 at 23:10
  • $\begingroup$ You're welcome. Everyone, who isn't explicitly explained that in their lectures, asks themselves this very question. This is the very first "weirdness" of QM you are introduced to. $\endgroup$ Commented Aug 7, 2021 at 23:12

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